Problem: $f(x)=x^2-2x+5$ What is the average rate of change of $f$ over the interval $-1\leq x \leq t$, in terms of $t$, where $t\neq -1$ ? Your answer must be fully expanded and simplified.
Solution: This is the formula for the average rate of change of a function $f$ over the interval $[a,b]$ : $\dfrac{f(b)-f(a)}{b-a}$ We can calculate that $f(-1)=8$. We are interested in the average rate of change of $f(x)=x^2-2x+5$ over the interval $-1\leq x \leq t$ : $\begin{aligned} &\phantom{=}\dfrac{f(t)-f(-1)}{(t)-(-1)} \\\\ &=\dfrac{t^2-2t+5-8}{t-(-1)} \\\\ &=\dfrac{t^2-2t-3}{t+1} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} \dfrac{t^2-2t-3}{t+1} &=\dfrac{(t-3)(t+1)}{t+1} \\\\ &=t-3\text{, for }t\neq -1 \end{aligned}$ Since we are given that $t\neq -1$, the average rate of change of the function is $t-3$. Notice that the average rate of change is calculated just like the slope of the secant line that intersects the graph of the function at the interval's endpoints.